# Waves: From Visuals to Equations

"Brian wouldn't rail against the dying of the light; he would simply capture the light in an equation." - from 'The Importance of being Interested'

Waves have always enchanted me - from sound waves to EM waves to the matter waves in quantum physics - they are everywhere. Here, we will try to construct an equation that describes a simple sinusoidal traveling wave and appreciate the beauty of our process in doing so.

Figure 1: Pulse and wave on a string: From Fundamentals of physics / Walker, Halliday, Resnick—10th edition.

First, let's try to understand what a wave is. Grab a string and tie one of its ends to a fixed object (like a wall) and hold the other end in your hand. Move that end of the string up and down once quickly. You will see a ‘pulse’ moving straight through the string. This is similar (not the same) to a stadium wave where successive groups of spectators briefly stand up and go back to the normal seated position (You can watch one here).

Now, if you moved your hand up and down continuously, you would see a wave travel along the string. What is interesting is that an element of the string only moves up and down (oscillates). It is the disturbance (wave) that moves through the string (Recall the Mexican wave example). Hereafter, I shall refer to the oscillating element of string as an oscillating particle.

Figure 2: The purple particle only oscillates up and down as the wave passes.

You might recognise that the shape of the wave is like the graph of sine and cosine functions. That fact is going to be very useful as we try to transform the wave into an equation. If you are completely new to trig functions or just need a review, check out live-help and trigonometry sessions on schoolhouse.

Before diving into the world of maths and equations, we must ask ourselves the question, what physical aspect of the wave are we trying to represent as an equation? Well, what do we see as a wave? We see its shape and its motion (think about that).

Let’s say that the wave moves in the y-x plane with positive y direction as straight up and positive x direction toward the right. With the convention set, we can now think of the shape of the wave like this - y represents the vertical displacement of the oscillating particle that is at x distance from the origin (Take a moment to absorb that.) It is also clear from Figure 2 that the vertical displacement of a given oscillating particle depends on the time - this is the motion aspect. Therefore, our aim will be to construct an equation that gives the vertical displacement of any oscillating particle as a function of its distance from the origin and time, i.e. y=f(x,t).

Consider two snapshots of the wave taken at two instants of time, say at t=0 and t=t'. In the diagram below, the dotted black curve represents the wave at t=0 and the solid black curve represents the wave at the later instant t'. We can see that the wave has moved by a distance = vt , where v is the speed of the wave and t is the time elapsed between the two snapshots.

Figure 3: The vertical displacement at x0 has moved to x in time t'

The disturbance (vertical displacement) that was initially at x0 has arrived at x at t=t' .To find out where the wave which is at x (at t') was at when t=0, we just need to subtract vt from x i.e. x0 =x-vt. As the wave travels from x0 to x, the vertical displacement doesn’t change. The vertical displacement can be associated with where the particle is in its cycle of oscillation.

For doing so, we will need to understand what wavelength is. Wavelength is the distance moved by the wave in the time a particle completes one oscillation. In the diagram below, as the purple particle oscillates, the wave moves to the right. The red arrow shows how far a particular trough has moved as the particle oscillates. The wave has covered a distance at the completion of one oscillation by the particle.

Figure 4: Snapshots of the waveform at successive instants of time (top to bottom)

To find out at how far into the cycle was the particle x0 at t=0, we can divide x0 by the wavelength. As lambda corresponds to a complete cycle, dividing x0 by will give us ‘what fraction of the cycle’ the particle was at.

Since the sine function takes its input as an angle, we will need to convert this fraction to an angle. As the oscillation repeats after each cycle, we can connect one cycle to an angle of 2pi. Therefore, ‘what fraction of a cycle’ can be converted to an angle as follows:

This quantity is called the phase of the particle. The phase of a particle is the angle that corresponds to the part of the cycle the particle is in. For example, a particle that is halfway into the cycle will have a phase of pi.

**Note about phase**: As the wave travels to the right, particles further to the right of a specific point will have the phase that it has now. Say, a crest moves to a different x - for it to remain a crest, it must have the same vertical displacement i.e. the same phase. We can therefore say that a constant phase point travels with the velocity of the waves. In the equation above, for it to remain constant as t increases, x must also increase. This indicates that the wave travels to the right. Isn’t that beautiful?

On taking the phase as the argument of sine function - you can connect the x coordinate of a particle to its vertical displacement.

Remember that sine function can only give you values that range from -1 to 1. But a wave can have any amplitude, so we shall multiply the sine function by a constant A to scale the vertical displacement to the required length.

We have successfully converted the x coordinate of any oscillating particle to its vertical displacement at any instant of time :)

Let’s change the notation of our equation to the one usually seen in textbooks. We will need to introduce two quantities for that:

- The propagation constant k=(2pi)/lambda is the change in phase per unit length - for two particles separated by lambda, the difference in phase is 2pi.
- The angular frequency omega=(2pi)/T is the angular analogue of frequency (the number of cycles completed in unit time). Therefore, omega can be understood as the phase change that occurs in a particle in unit time i.e. 2pi phase change in T (where T is the time taken to complete an oscillation)

On substituting v as T (speed is distance per time), we get:

Let’s simplify and use the equations for k and .

There’s just one more tweak needed to generalise this equation. This equation assumes that the initial waveform is in the shape of a sine wave.

In general, the phase of the particle at origin at t=0 need not be 0. Let’s assume the particle at origin was at a quarter of the cycle at t=0, which corresponds to a phase of pi/2. Therefore, at t=0 and x=0, the phase should be pi/2. To account for this, we will need to add pi/2 to the argument of the sine function, changing our wave equation to:

To generalise, if the particle at origin has a phase of phi initially, the displacement equation is:

This is the wave displacement equation for a wave travelling in the +x direction. I hope you enjoyed the process of constructing the equation.

**Challenge: Construct the displacement equation for a wave travelling to the left. What does the phase remaining constant mean here?**

Let’s always learn to look beyond the Greek letters & mathy symbols to understand the true meaning and beauty of every equation. :)

**References**

- Fundamentals of physics / Jearl Walker, David Halliday, Robert Resnick—10th edition.
- https://www.physicsforums.com/threads/phase-of-a-wave-my-interpretation.1046426/#post-6811347
- https://ncert.nic.in/textbook.php?keph2=7-7
- https://physics.stackexchange.com/questions/382774/phase-of-wave-is-constant

Thank you to Hafsah M for editing this article!

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